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144 ANALYTICAL MECHANICS 3. Find the center of mass of a semicircular lamina. Selecting the coordinates and the element of mass as shown in Fig.' we have dm = cr-pd6 ğdp, nay crp dp d6 _ >_____________ V~ C'C'rpdpdB Ğ/0 J 0 O FIG. 76. = 0. PROBLEMS. Find the center of mass of the lamina bounded by the following curves: (1) y = mx, y mx, and y a. (2) y = asi&x, y = 0, x = 0, and x = TT. (3) yz = ax and #2 = by. (4) x2 + 7/2 = a2, Z = 0, and y = 0. (5) bV + a22/2 = a262, x = 0, and a/ = 0. (6) r=a(l+cos0). (7) r = a, 0 = 0, and 6 = 0. (8) r = a, r = b, 6 = 0, anc 7T V 116. Center of Mass of a Homogeneous Solid of Revolution. Let Fig. 77 represent any solid . obtained by revolving a plane curve about the x-axis. Then the center of mass lies on the axis of revolution. The position of the center of mass is found most conveniently when the element of mass is a thin slice obtained by two transverse sections.. The expression for the mass of such an element is dm = r wy2 - dx, where r is the density of the solid, y the radius of the slice,, and dx its thickness.